788
SECTION 16.1
CHAPTER 16
SECTION 16.1
1. ∇f = (6x − y) i + (1 − x) j 2. ∇f = (2Ax + By)i + (Bx + 2Cy)j 3. ∇f = exy[ (xy + 1) i + x2j]
4. ∇f = 1
(x2+ y2)2[(y
2− x2+ 2xy)i + (y2− x2− 2xy)j]
5. ∇f =2y2sin(x2+ 1) + 4x2y2cos(x2+ 1)i + 4xy sin(x2+ 1) j
6. ∇f = 2x
x2+ y2i +
2y x2+ y2j
7. ∇f = (ex−y+ ey−x) i + (−ex−y− ey−x) j = (ex−y+ ey−x)(i− j)
8. ∇f = AD− BC (Cx + Dy)2[ yi− xj ] 9. ∇f = (z2+ 2xy) i + (x2+ 2yz) j + (y2+ 2zx) k 10. ∇f = x x2+ y2+ z2i + y x2+ y2+ z2j + z x2+ y2+ z2k 11. ∇f = e−z(2xy i + x2j− x2y k) 12. ∇f = xyz x + y + z + yz ln(x + y + z) i + xyz x + y + z+ xz ln(x + y + z) j + xyz x + y + z + xy ln(x + y + z) k
13. ∇f = ex+2ycosz2+ 1i + 2ex+2ycosz2+ 1j− 2zex+2ysinz2+ 1k
14. ∇f = eyz2/x3 −3yz2 x4 i + z2 x3j + 2yz x3 k 15. ∇f = 2y cos(2xy) + 2 x i + 2x cos(2xy) j +1 zk 16. ∇f = 2xy z − 3z 4 i +x 2 z j− x2y z2 + 12xz 3 k
17. ∇f = (4x − 3y) i + (8y − 3x) j; at (2, 3), ∇f = −i + 18j 18. ∇f = 1 (x− y)2(−2yi + 2xj), ∇f(3, 1) = − 1 2i + 3 2j 19. ∇f = 2x x2+ y2i + 2y x2+ y2j; at (2, 1), ∇f = 4 5i + 2 5j
20. ∇f = tan−1(y/x)− xy x2+ y2 i + x2 x2+ y2 j, ∇f(1, 1) = π 4 − 1 2 i +1 2j
21. ∇f = (sin xy + xy cos xy) i + x2 cos xy j; at (1, π/2), ∇f = i
22. ∇f = e−(x2+y2)[(y− 2x2y)i + (x− 2xy2)j], ∇f(1, −1) = e−2(i− j)
23. ∇f = −e−xsin (z + 2y) i + 2e−xcos (z + 2y) j + e−xcos (z + 2y) k; at (0, π/4, π/4), ∇f = −12√2 (i + 2j + k)
24. ∇f = cos πzi − cos πzj − π(x − y) sin πzk, ∇f
1, 0,1 2 =−πk 25. ∇f = i − y y2+ z2j− z y2+ z2k; at (2,−3, 4), ∇f = i + 3 5j− 4 5k
26. ∇f = − sin(xyz2)(yz2i + xz2j + 2xyzk), ∇f
π,1 4,−1 =− √ 2 2 1 4i + π j− π 2 k 27. (a) ∇f(0, 2) = 4 i (b) ∇f14π,16π= −1 −−1 + √ 3 2√2 i + −1 2 − −1 +√ √3 2 j (c) ∇f(1, e) = (1 − 2e) i − 2 j 28. (a) ∇f(1, 2, −3) = 1 8√2i + 1 2√2j− 27 8√2k (b) ∇f(1, −2, 3) = − 5 18i + 1 9j + 1 18k (c) ∇f(1, e2, π/6) = √ 3 2 i + π 12e2j + k
29. For the function f (x, y) = 3x2− xy + y, we have
f (x + h)− f(x) = f(x + h1, y + h2)− f(x, y) = 3(x + h1)2− (x + h1)(y + h2) + (y + h2)− 3x2− xy + y = [(6x− y) i + (1 − x) j] · (h1i + h2j) + 3h21− h1h2 = [(6x− y) i + (1 − x) j] · h + 3h2 1− h1h2 The remainder g(h) = 3h2 1− h1h2= (3h1i− h1j) · (h1i + h2j) , and |g(h)| h = 3h1i− h1j · h · cos θ h ≤ 3h1i− h1j
Since 3h1i− h1j → 0 as h → 0 it follows that
∇f = (6x − y) i + (1 − x) j
30. f (x + h)− f(x) = [(x + 2y) i + (2x + 2y) j] · [h1i + h2j] + 12h21+ 2h1h2+ h22;
g(h) =12h2
790
SECTION 16.1
31. For the function f (x, y, z) = x2y + y2z + z2x, we have
f (x + h)− f(x) = f(x + h1, y + h2, z + h3)− f(x, y, z) = (x + h1)2(y + h2) + (y + h2)2(z + h3) + (z + h3)2(x + h1)− x2y + y2z + z2x =2xy + z2h1+ 2yz + x2h2+ 2xz + y2h3+ (2xh2+ yh1+ h1h2) h1+ (2yh3+ zh2+ h2h3) h2+ (2zh1+ xh3+ h1h3) h3 =2xy + z2i +2yz + x2j +2xz + y2k · h + g(h) · h, where g(h) = (2xh2+ yh1+ h1h2) i + (2yh3+ zh2+ h2h3) j + (2zh1+ xh3+ h1h3) k Since |g(h)| h → 0 as h → 0 it follows that ∇f =2xy + z2i +2yz + x2j +2xz + y2k 32. f (x + h)− f(x) = (2xy + 2h2x + h1y) i + 2x2j + 1 z(z + h3) k · (h1i + h2j + h3k) + h21; g(h) = h2
1h2 is o(h) and∇f = 4xy i = 2x2j +z12k.
33. ∇f = F(x, y) = 2xy i +1 + x2j ⇒ ∂f
∂x = 2xy ⇒ f(x, y) = x
2y + g(y) for some function g.
Now, ∂f
∂y = x
2+ g(y) = 1 + x2 ⇒ g(y) = 1 ⇒ g(y) = y + C, C a constant. Thus, f (x, y) = x2y + y + C
34. ∇f = (2xy + x)i + (x2+ y)j =⇒ fx= 2xy + x =⇒ f(x, y) = x2y + 1 2x
2+ g(y)
Now, fy= x2+ g(y) = x2+ y =⇒ g(y) = y =⇒ g(y) = 12y2+ C Thus, f (x, y) = x2y +1
2x2+ 1 2y2+ C
35. ∇f = F(x, y) = (x + sin y) i + (x cos y − 2y) j ⇒ ∂f
∂x = x + sin y ⇒ f(x, y) =
1 2x
2+ x sin y + g(y)
for some function g.
Now, ∂f
∂y = x cos y + g
(y) = x cos y− 2y ⇒ g(y) =−2y ⇒ g(y) = −y2+ C, C a constant.
Thus, f (x, y) = 12x2+ x sin y− y2+ C.
36. ∇f = yzi + (xz + 2yz)j + (xy + y2)k =⇒ fx= yz =⇒ f(x, y, z) = xyz + g(y, z).
fy= xz + gy = xz + 2yz =⇒ gy= 2yz =⇒ g(y, z) = y2z + h(z) =⇒ f(x, y, z) = xyz + y2z + h(z). fx= xy + y2+ h(z) = xy + y2 =⇒ h(z) = 0 =⇒ h(z) = C. Thus, f (x, y, z) = xyz + y2z + C. 37. With r = (x2+ y2+ z2)1/2 we have ∂r ∂x = x r, ∂r ∂y = y r, ∂r ∂z = z r. (a) ∇(ln r) = ∂ ∂x(ln r) i + ∂ ∂y(ln r) j + ∂ ∂z(ln r)k = 1 r ∂r ∂xi + 1 r ∂r ∂yj + 1 r ∂r ∂zk = x r2i + y r2j + z r2k = r r2
(b) ∇(sin r) = ∂ ∂x(sin r) i + ∂ ∂y(sin r) j + ∂ ∂z(sin r)k = cos r ∂r ∂xi + cos r ∂r ∂yj + cos r ∂r ∂zk = (cos r)x ri + (cos r) y rj + (cos r) z rk = cos r r r (c)∇er= er r
r [ same method as in (a) and (b) ]
38. With rn = (x2+ y2+ z2)n/2 we have ∂rn ∂x = n 2(x 2+ y2+ z2)(n/2)−1(2x) = n(x2+ y2+ z2)(n−2)/2x = nrn−2x. Similarly ∂rn ∂y = nr n−2y and ∂rn ∂z = nr n−2z. Therefore ∇rn= nrn−2xi + nrn−2yj + nrn−2zk = nrn−2(xi + yj + zk) = nrn−2r 39. (a)∇f = 2x i + 2y j = 0 =⇒ x = y = 0; ∇f = 0 at (0, 0).
(b) (c) f has an absolute minimum at (0, 0)
40. (a) ∇f = −1
4− x2− y2(xi + yj) = 0 at (0, 0) (b)
792
SECTION 16.2
41. (a) Let c = c1i + c2j + c3k. First, we take h = hi. Since c· h is o(h),
0 = lim h→0 c· h h = limh→0 c1h h = c1. Similarly, c2= 0 and c3= 0.
(b) (y− z) · h = [f(x + h) − f(x) − z · h] + [y · h − f(x + h) + f(x) ] = o(h) + o(h) = o(h), so that, by part (a), y− z = 0.
42. lim h→0g(h) = limh→0 hg(h) h = lim h→0h lim h→0 g(h) h = (0)(0) = (0).
43. (a) In Section 15.6 we showed that f was not continuous at (0, 0). It is therefore not differentiable
at (0, 0).
(b) For (x, y)= (0, 0), ∂f ∂x =
2y(y2− x2)
(x2+ y2)2 . As (x, y) tends to (0, 0) along the positive y-axis,
∂f ∂x = 2y3 y4 = 2 y tends to∞. SECTION 16.2 1. ∇f = 2xi + 6yj, ∇f(1, 1) = 2i + 6j, u = 12√2 (i− j), fu(1, 1) =∇f(1, 1) · u = −2 √ 2
2. ∇f = [1 + cos(x + y)]i + cos(x + y)j, ∇f(0, 0) = 2i + j, u = √1
5(2i + j), fu(0, 0) =∇f(0, 0) · u = √ 5 3. ∇f = (ey− yex) i + (xey− ex) j, ∇f(1, 0) = i + (1 − e)j, u = 1 5(3i + 4j), fu(1, 0) =∇f(1, 0) · u = 1 5(7− 4e) 4. ∇f = 1 (x− y)2(−2yi + 2xj), ∇f(1, 0) = 2j, u = 1 2(i− √ 3j), fu(1, 0) =∇f(1, 0) · u = − √ 3 5. ∇f = (a− b)y (x + y)2i + (b− a)x (x + y)2j, ∇f(1, 1) = a− b 4 (i− j), u = 1 2 √ 2 (i− j), fu(1, 1) =∇f(1, 1) · u = 1 4 √ 2 (a− b) 6. ∇f = 1 (cx + dy)2[(d− c)yi + (c − d)xj] , ∇f(1, 1) = d− c (c + d)2(i− j), u = 1 √ c2+ d2(ci− dj), fu(1, 1) =∇f(1, 1) · u = d− c (c + d)√c2+ d2 7. ∇f = 2x x2+ y2i + 2y x2+ y2j, ∇f(0, 1) = 2 j, u = 1 √ 65(8 i + j), fu(0, 1) =∇f(0, 1) · u = 2 √ 65
8. ∇f = 2xyi + (x2+ sec2y)j, ∇f(−1,π 4) =− π 2i + 3j, u = 1 √ 5(i− 2j) fu −1,π 4 =∇f −1,π 4 · u = −√1 5 π 2 + 6 9. ∇f = (y + z)i + (x + z)j + (y + x)k, ∇f(1, −1, 1) = 2j, u =16√6 (i + 2j + k), fu (1,−1, 1) = ∇f(1, −1, 1) · u = 2 3 √ 6 10. ∇f = (z2+ 2xy)i + (x2+ 2yz)j + (y2+ 2zx)k, ∇f(1, 0, 1) = i + j + 2k, u = √1 10(3j− k) fu(1, 0, 1) =∇f(1, 0, 1) · u = √ 10 10 11. ∇f = 2x + y2+ z3 i + 2yj + 3z2k, ∇f(1, −1, 1) = 6(i − 2j + 3k), u =1 2 √ 2 (i + j), fu(1,−1, 1) = ∇f(1, −1, 1) · u = −3 √ 2
12. ∇f = (2Ax + Byz)i + (Bxz + 2Cy)j + Bxyk, ∇f(1, 2, 1) = 2(A + B)i + (B + 4C)j + 2Bk
u = √ 1 A2+ B2+ C2(Ai + Bj + Ck); f u(1, 2, 1) =∇f(1, 2, 1) · u = 2A2+ B2+ 2AB + 6BC √ A2+ B2+ C2 13. ∇f = tan−1(y + z) i + x 1 + (y + z)2j + x 1 + (y + z)2k, ∇f(1, 0, 1) = π 4 i + 1 2j + 1 2k, u = √1 3(i + j− k), f u(1, 0, 1) =∇f(1, 0, 1) · u = π 4√3 = √ 3 12 π
14. ∇f = (y2cos z− 2πyz2cos πx + 6zx)i + (2xy cos z− 2z2sin πx)j + (−xy2sin z− 4yz sin πx + 3x2)k
∇f(0, −1, π) = (2π3− 1)i; u = 1 3(2i− j + 2k), f u(0,−1, π) = ∇f(0, −1, π) · u = 2 3(2π 3− 1). 15. ∇f = x x2+ y2i + y x2+ y2j, u = 1 x2+ y2 (−xi − yj) , f u(x, y) =∇f · u = − 1 x2+ y2 16. ∇f = exy(y2+ xy3− y3)i + (x− 1)(2y + xy2)j, ∇f(0, 1) = −2j u = √1 5(−i + 2j), f u(0, 1) =∇f(0, 1) · u = − 4 5 √ 5
17. ∇f = (2Ax + 2By) i + (2Bx + 2Cy) j, ∇f(a, b) = (2aA + 2bB)i + (2aB + 2bC) j
(a) u = 12√2 (−i + j), fu(a, b) =∇f(a, b) · u =
√ 2 [a(B− A) + b(C − B)] (b) u = 12√2 (i− j), fu(a, b) =∇f(a, b) · u = √ 2 [a(A− B) + b(B − C)] 18. ∇f = z xi− z yj + ln x y k, ∇f(1, 1, 2) = 2i − 2j u = √1 3(i + j− k); f u(1, 1, 2) =∇f(1, 1, 2) · u = 0
794
SECTION 16.2
19. ∇f = ey2−z2 (i + 2xyj− 2xzk), ∇f(1, 2, −2) = i + 4j + 4k, r(t) = i− 2 sin (t − 1) j − 2et−1k, at (1, 2,−2) t = 1, r(1) = i− 2k, u = 15√5 (i− 2k), fu(1, 2,−2) = ∇f(1, 2, −2) · u = − 7 5 √ 5 20. ∇f = 2xi + zj + yk, ∇f(1, −3, 2) = 2i + 2j − 3k Direction: r(−1) = −2i + 3j − 3k, u =√1 22(−2i + 3j − 3k), f u(1,−3, 2) = ∇f(1, −3, 2) · u = 1 2 √ 2221. ∇f = (2x + 2yz) i +2xz− z2j + (2xy− 2yz) k, ∇f(1, 1, 2) = 6 i − 2 k
The vectors v =±(2 i + j − 3 k) are direction vectors for the given line; u = ±
1 √
14[2 i + j− 3 k] are corresponding unit vectors; fu(1, 1, 2) =∇f(1, 1, 2) · (±u) = ±
18 √
14
22. ∇f = ex(cos πyzi− πz sin πyzj − πy sin πyzk), ∇f(0, 1,1 2) =−
π 2 j− π k The vectors v =±(2 i + 3 j + 5 k) are direction vectors for the line; u = ±
1 √
38[2 i + 3 j + 5 k] are corresponding unit vectors; fu(0, 1,12) =∇f(0, 1,
1 2)· (±u) = ∓ 13π 2√38 23. ∇f = 2y2e2xi + 2ye2xj, ∇f(0, 1) = 2 i + 2 j, ∇f = 2√2, ∇f ∇f = 1 √ 2(i + j) f increases most rapidly in the direction u = √1
2(i + j); the rate of change is 2 √
2. f decreases most rapidly in the direction v =−√1
2(i + j); the rate of change is−2 √
2.
24. ∇f = [1 + cos(x + 2y)]i + 2 cos(x + 2y)j, ∇f(0, 0) = 2i + 2j
Fastest increase in direction u = √1
2(i + j), rate of change ∇f(0, 0) = 2 √
2 Fastest decrease in direction v =−√1
2(i + j), rate of change −2 √ 2 25. ∇f = x x2+ y2+ z2i + y x2+ y2+ z2j + z x2+ y2+ z2k, ∇f(1, −2, 1) = √1 6(i− 2 j + k), ∇f = 1 f increases most rapidly in the direction u = √1
6(i− 2 j + k); the rate of change is 1. f decreases most rapidly in the direction v =−√1
6(i− 2 j + k); the rate of change is −1.
26. ∇f = (2xzey+ z2)i + x2zeyj + (x2ey+ 2xz)k, ∇f(1, ln 2, 2) = 12i + 4j + 6k Fastest increase in direction u = 1
7(6i + 2j + 3k), rate of change ∇f(1, ln 2, 2) = 14 Fastest decrease in direction v =−1
7(6i + 2j + 3k), rate of change −14
27. ∇f = f(x0) i. If f(x0)= 0, the gradient points in the direction in which f increases: to the right
28. 0; the vector c = ∂f
∂y(x0, y0)i− ∂f
∂x(x0, y0)j is perpendicular to the gradient∇f(x0, y0) and points along the level curve of f at (x0, y0).
29. (a) lim h→0 f (h, 0)− f(0, 0) h = limh→0 √ h2 h = limh→0 |h|
h does not exist (b) no; by Theorem 16.2.5 f cannot be differentiable at (0, 0)
30. (a) g(x + h)o(h) h = g(x + h) o(h) h → g(x)(0) = 0 (b) |[g(x + h) − g(x)]∇f(x) · h| h ≤ [g(x + h) − g(x)]∇f(x)h h by Schwarz’s inequality =|g(x + h) − g(x)| · ∇f(x) → 0 31. ∇λ(x, y) = −8 3xi− 6yj (a) ∇λ(1, −1) = −8 3i = 6j, u = −∇λ(1, −1) ∇λ(1, −1)= 8 3i− 6j 2 3 √ 97 , λ u(1,−1) = ∇λ(1, −1) · u = − 2 3 √ 97 (b) u = i, λu(1, 2) =∇λ(1, 2) · u = −8 3i− 12j · i = −8 3 (c) u = 12√2 (i + j), λu(2, 2) =∇λ(2, 2) · u = −16 3 i− 12 j · 1 2 √ 2 (i + j)=−263√2
32. ∇I = −4xi − 2yj. We want the curve r(t) = x(t)i + y(t)j which begins at (−2, 1) and has tangent
vector r(t) in the direction∇I. We can satisfy these conditions by setting x(t) =−4x(t), x(0) = −2; y(t) =−2y(t), y(0) = 1. These equations imply that
x(t) =−2e−4t, y(t) = e−2t.
Eliminating the parameter, we get x =−2y2; the particle will follow the parabolic path x =−2y2
toward the origin.
33. (a) The projection of the path onto the xy-plane is the curve
C : r(t) = x(t)i + y(t)j
which begins at (1, 1) and at each point has its tangent vector in the direction of−∇f. Since
∇f = 2xi + 6yj,
we have the initial-value problems
x(t) =−2x(t), x(0) = 1 and y(t) =−6y(t), y(0) = 1. From Theorem 7.6.1 we find that
x(t) = e−2t and y(t) = e−6t.
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SECTION 16.2
(b) Here
x(t) =−2x(t), x(0) = 1 and y(t) =−6y(t), y(0) = −2 so that
x(t) = e−2t and y(t) =−2e−6t.
Eliminating the parameter t, we find that the projection of the path onto the xy-plane is the curve y =−2x3 from (1,−2) to (0, 0).
34. z = f (x, y) = 1
2x
2− y2; ∇f = xi − 2yj, so we choose the projection r(t) = x(t)i + y(t)j of the path
onto the xy-plane such that x(t) = x(t), y(t) =−2y(t)
(a) With initial point (−1, 1, −12), we get x(t) =−et, y(t) = e−2t, or y = 1 x2
from (−1, 1), in the direction of decreasing x.
(b) With initial point (1, 0,12), we get x(t) = et, y(t) = 0, or the x-axis from (1, 0), in the direction of increasing x.
35. The projection of the path onto the xy-plane is the curve
C : r(t) = x(t)i + y(t)j
which begins at (a, b) and at each point has its tangent vector in the direction of −∇f = −2a2xi + 2b2yj. We can satisfy these conditions by setting
x(t) =−2a2x(t), x(0) = a2 and y(t) =−2b2y(t), y(0) = b so that
x(t) = ae−2a2t and y(t) = be−2b2t . Since x a b2 = e−2a2t b2 = y b a2 , C is the curve (b)a2xb2 = (a)b2ya2 from (a, b) to (0, 0).
36. The particle must go in he direction−∇T = −eycos xi− eysin xj, so we set x(t) =−ey(t)cos x(t), y(t) =−ey(t)sin x(t). Dividing, we have y(t)
x(t) =
sin x(t) cos x(t), or
dy
dx = tan x. With initial point (0, 0), we get y = ln| sec x|, in the direction of decreasing x (since x(0) < 0).
37. We want the curve
C : r(t) = x(t)i + y(t)j
which begins at (π/4, 0) and at each point has its tangent vector in the direction of
∇T = −√2 e−ysin x i−√2 e−ycos x j. From
we obtain dy dx = y(t) x(t) = cot x so that y = ln| sin x| + C.
Since y = 0 when x = π/4, we get C = ln√2 and y = ln|√2 sin x|. As ∇T (π/4, 0) = −i − j, the curve y = ln|√2 sin x| is followed in the direction of decreasing x.
38. ∇z = (1 − 2x)i + (2 − 6y)j, so the projection of the path onto the xy-plane satisfies x(t) = 1− 2x(t), y(t) = 2− 6y(t), or dy
dx = 2− 6y
1− 2x. With initial point (0, 0), this gives the curve 3y = (2x− 1)3+ 1, in the direction of increasing x.
39. (a) lim h→0 f2 + h, (2 + h)2− f(2, 4) h = limh→0 3(2 + h)2+ (2 + h)2− 16 h = lim h→04 4h + h2 h = lim h→04(4 + h) = 16 (b) lim h→0 f h + 8 4 , 4 + h − f(2, 4) h = limh→0 3 h + 8 4 2 + (4 + h)− 16 h = lim h→0 3 16h2+ 3h + 12 + 4 + h− 16 h = lim h→0 3 16h + 4 = 4 (c) u = 1 17 √ 17 (i + 4j), ∇f(2, 4) = 12i + j; fu(2, 4) =∇f(2, 4) · u = 1617 √ 17
(d) The limits computed in (a) and (b) are not directional derivatives. In (a) and (b) we have, in essence, computed∇f(2, 4) · r0 taking r0= i + 4j in (a) and r0= 14i + j in (b). In neither case
is r0a unit vector.
40. ∇f = −GMm
(x2+ y2+ z2)3/2(xi + yj + zk) =
−GMm r3 r
41. (a) u = cos θ i + sin θ j, ∇f(x, y) = ∂f
∂xi + ∂f ∂y j; fu(x, y) =∇f · u = ∂f ∂xi + ∂f ∂yj · (cos θ i + sin θ j) =∂f ∂x cos θ + ∂f ∂y sin θ (b) ∇f =3x2+ 2y− y2i + (2x− 2xy) j, ∇f(−1, 2) = 3 i + 2 j fu(−1, 2) = 3 cos(2π/3) + 2 sin(2π/3) = 2√3− 3 2
798
SECTION 16.3
42. fu(x, y) = ∂f ∂xcos 5π 4 + ∂f ∂ysin 5π 4 = 2xe 2y − √ 2 2 + 2x2e2y − √ 2 2 =−√2xe2y(1 + x) fu(2, ln 2) =− √ 2· 2 · e2 ln 2(1 + 2) =−24√2 43. ∇(fg) = ∂(f g) ∂x i + ∂(f g) ∂y j + ∂(f g) ∂z k = f ∂g ∂x + g ∂f ∂x i + f ∂g ∂y+ g ∂f ∂y j + f ∂g ∂z + g ∂f ∂z k = f ∂g ∂xi + ∂g ∂yj + ∂g ∂zk + g ∂f ∂xi + ∂f ∂y j + ∂f ∂zk = f∇g + g ∇f 44. ∇ f g = ∂ ∂x f g i + ∂ ∂y f g j + ∂ ∂z f g k = ∂f ∂xg− f ∂g ∂x g2 i + ∂f ∂yg− f ∂g ∂y g2 j + ∂f ∂zg− f ∂g ∂z g2 k = g(x)∇f(x) − f(x)∇g(x) g2(x) 45. ∇fn= ∂fn ∂x i + ∂fn ∂y j + ∂fn ∂z k = nf n−1∂f ∂xi + nf n−1∂f ∂y j + nf n−1∂f ∂z k = nf n−1∇f SECTION 16.3 1. f (b) = f (1, 3) =−2; f(a) = f(0, 1) = 0; f(b) − f(a) = −2 ∇f =3x2− yi− x j; b − a = i + 2 j and ∇f · (b − a) = 3x2− y − 2xThe line segment joining a and b is parametrized by
x = t, y = 1 + 2t, 0≤ t ≤ 1 Thus, we need to solve the equation
3t2− (1 + 2t) − 2t = −2, which is the same as 3t2− 4t + 1 = 0, 0 ≤ t ≤ 1 The solutions are: t = 1
3, t = 1. Thus, c = ( 1 3,
5
3) satisfies the equation.
Note that the endpoint b also satisfies the equation.
2. ∇f = 4zi − 2yj + (4x + 2z)k, f(a) = f(0, 1, 1) = 0, f(b) = f(1, 3, 2) = 3 b− a = i + 2j + k, so we want (x, y, z) such that
∇f · (b − a) = 4z − 4y + 4x + 2z = 6z − 4y + 4x = f(b) − f(a) = 3
Parametrizing the line segment from a to b by x(t) = t, y(t) = 1 + 2t, z(t) = 1 + t, we get t = 12, or c = (12, 2,32)
3. (a) f (x, y, z) = a1x + a2y + a3z + C (b) f (x, y, z) = g(x, y, z) + a1x + a2y + a3z + C
5. (a) U is not connected
(b) (i) g(x) = f (x)− 1 (ii) g(x) =−f(x)
6. By the mean-value theorem
f (x1)− f(x2) =∇f(c) · (x1− x2)
for some point c on the line segment x1x2. Since Ω is convex, c is in Ω. Thus
|f(x1)− f(x2)| = |∇f(c) · (x1− x2)| ≤ ∇f(c)x1− x2 ≤ Mx1− x2.
by Schwarz’s inequality ∧
7. ∇f = 2xyi + x2j;
∇f(r(t)) · r(t) =2i + e2tj · (eti− e−tj) = et
8. ∇f = i − j; ∇f(r(t)) · r(t) = (i− j) · (ai − ab sin atj) = a(1 + b sin at)
9. ∇f = −2x 1 + (y2− x2)2i + 2y 1 + (y2− x2)2j, ∇f(r(t)) = −2 sin t 1 + cos22ti + 2 cos t 1 + cos22tj ∇f(r(t)) · r(t) = −2 sin t 1 + cos22ti + 2 cos t 1 + cos22tj
· (cos t i − sin t j) =−4 sin t cos t
1 + cos22t = −2 sin 2t 1 + cos22t 10. ∇f = 1 2x2+ y3(4xi + 3y 2j) ∇f(r(t)) · r(t) = 1 2e4t+ t(4e 2ti + 3t2/3j)· (2e2ti +1 3t −2/3j) = 8e4t+ 1 2e4t+ t 11. ∇f = (ey− ye−x) i + (xey+ e−x) j; ∇f(r(t)) = (tt− ln t) i + ttln t + 1 t j ∇f(r(t)) · r(t) = (tt− ln t) i + ttln t + 1 t j · 1 ti + [1 + ln t] j = tt 1 t + ln t + [ln t] 2 +1 t 12. ∇f = 2 x2+ y2+ z2(xi + yj + zk) ∇f(r(t)) · r(t) = 2 1 + e4t(sin ti + cos tj + e 2tk)· (cos ti − sin tj + 2e2tk) = 4e4t 1 + e4t 13. ∇f = yi + (x − z)j − yk; ∇f(r(t)) · r(t) =t2i +t− t3j− t2k · i + 2tj + 3t2k= 3t2− 5t4 14. ∇f = 2x i + 2y j
∇f(r(t)) · r(t) = (2a cos ωti + 2b sin ωtj)· (−ωa sin ωti + ωb cos ωtj + bωk) = 2ω(b2− a2) sin ωt cos ωt
15. ∇f = 2xi + 2yj + k;
∇f(r(t)) · r(t) = (2a cos ωt i + 2b sin ωt j + k)· (−aω sin ωt i + bω cos ωt j + bωk) = 2ωb2− a2sin ωt cos ωt + bω
800
SECTION 16.3
16. ∇f = y2cos(x + z)i + 2y sin(x + z)j + y2cos(x + z)k
∇f(r(t)) · r(t)
= [cos2t cos(2t + t3)i + 2 cos t sin(2t + t3)j + cos2t cos(2t + t3)k]· (2i − sin tj + 3t2k)
= cos t[(2 + 3t2) cos t cos(2t + t3)− 2 sin t sin(2t + t3)]
17. du dt = ∂u ∂x dx dt + ∂u ∂y dy
dt = (2x− 3y)(− sin t) + (4y − 3x)(cos t) = 2 cos t sin t + 3 sin2t− 3 cos2t = sin 2t− 3 cos 2t
18. du dt = ∂u ∂x· dx dt + ∂u ∂y · dy dt = 1 + 2 y x 3t2+ 2 x y − 3 −1 t2 = 1 + 2 t2 3t2+ (2t2− 3) −1 t2 = 3t2+ 4 + 3 t2 19. du dt = ∂u ∂x dx dt + ∂u ∂y dy dt = (exsin y + eycos x)1 2 + (excos y + eysin x) (2) = et/21 2sin 2t + 2 cos 2t + e2t1 2cos 1 2t + 2 sin 1 2t 20. du dt = ∂u ∂x· dx dt + ∂u ∂y · dy
dt = (4x− y)(−2 sin 2t) + (2y − x) cos t = 2 sin 2t(sin t− 4 cos 2t) + cos t(2 sin t − cos 2t)
21. du dt = ∂u ∂x dx dt + ∂u ∂y dy dt = (e xsin y) (2t) + (excos y) (π) = et2 [2t sin(πt) + π cos(πt)] 22. du dt = ∂u ∂x· dx dt + ∂u ∂y · dy dt + ∂u ∂z · dz dt =− z x2t + z y 1 2√t+ ln y x et(1 + t) =−2t 2et t2+ 1+ et 2 + ln √ t t2+ 1 et(1 + t) 23. du dt = ∂u ∂x dx dt + ∂u ∂y dy dt + ∂u ∂z dz dt = (y + z)(2t) + (x + z)(1− 2t) + (y + x)(2t − 2) = (1− t)(2t) + (2t2− 2t + 1)(1 − 2t) + t(2t − 2) = 1− 4t + 6t2− 4t3 24. du dt = ∂u ∂x· dx dt + ∂u ∂y · dy dt + ∂u ∂z · dz
dt = (sin πy + πz sin πx)2t + πx cos πy(−1) − cos πx(−2t) = 2tsin[π(1− t)] + π(1 − t2) sin(πt2)− πt2cos[π(1− t)] + 2t cos(πt2)
25. V = 1 3πr 2h, dV dt = ∂V ∂r dr dt + ∂V ∂h dh dt = 2 3πrh dr dt + 1 3πr 2 dh dt.
At the given instant, dV dt = 2 3π(280)(3) + 1 3π(196)(−2) = 1288 3 π. The volume is increasing at the rate of 1288
3 π in. 3/ sec . 26. v = πr2h, dv dt = ∂v ∂r· dr dt + ∂v ∂h · dh dt = 2πrh dr dt + πr 2dh dt dr dt =−2, dh dt = 3, r = 13, h = 18 =⇒ dv
dt =−429π : decreasing at the rate of 429π cm3/sec. 27. A = 12xy sin θ; dA dt = ∂A ∂x dx dt + ∂A ∂y dy dt + ∂A ∂θ dθ dt = 1 2 (y sin θ)dx dt + (x sin θ) dy dt + (xy cos θ) dθ dt . At the given instant
dA dt =
1
2[(2 sin 1) (0.25) + (1.5 sin 1) (0.25)− (2(1.5) cos 1) (0.1)] ∼= 0.2871 ft
2/s ∼ = 41.34 in2/s 28. dz dt = 2x dx dt + y 2 dy dt. But x 2+ y2= 13 =⇒ 2xdx dt + 2y dy dt = 0 =⇒ dy dt =− x y dx dt =⇒ dz dt = 2x dx dt + y 2 −x y dx dt =3x 2 dx
dt = 15. z is increasing 15 centimeters per second
29. ∂u ∂s = ∂u ∂x ∂x ∂s+ ∂u ∂y ∂y ∂s = (2x− y)(cos t) + (−x)(t cos s) = 2s cos2t− t sin s cos t − st cos s cos t
∂u ∂t = ∂u ∂x ∂x ∂t + ∂u ∂y ∂y
∂t = (2x− y)(−s sin t) + (−x)(sin s) =−2s2cos t sin t + st sin s sin t− s cos t sin s
30. ∂u ∂s = ∂u ∂x · ∂x ∂s+ ∂u ∂y · ∂y ∂s
= [cos(x− y) − sin(x + y)]t + [− cos(x − y) − sin(x + y)]2s = (t− 2s) cos(st − s2+ t2)− (t + 2s) sin(st + s2− t2) ∂u ∂t = ∂u ∂x · ∂x ∂t + ∂u ∂y · ∂y ∂t
= [cos(x− y) − sin(x + y)]s + [− cos(x − y) − sin(x + y)](−2t) = (s + 2t) cos(st− s2+ t2)− (s − 2t) sin(st + s2− t2) 31. ∂u ∂s = ∂u ∂x ∂x ∂s + ∂u ∂y ∂y ∂s = (2x tan y)(2st) + x2sec2y(1) = 4s3t2tans + t2+ s4t2sec2s + t2 ∂u ∂t = ∂u ∂x ∂x ∂t + ∂u ∂y ∂y ∂t = (2x tan y) s2+x2sec2y(2t) = 2s4t tans + t2+ 2s4t3sec2s + t2
802
SECTION 16.3
32. ∂u ∂s = ∂u ∂x· ∂x ∂s + ∂u ∂y · ∂y ∂s+ ∂u ∂z · ∂z ∂s= z2y sec xy tan xy(2t) + z2x sec xy tan xy + 2z sec xy(2st)
= sec[2st(s− t2)]2s4t3(s− t2) tan[2st(s− t2)] + 2s3t2tan[2st(s− t2)] + 4s3t2 ∂u
∂t = z
2y sec xy tan xy(2s) + z2x sec xy tan xy(−2t) + 2z sec xy(s2)
= sec[2st(s− t2)]2s5t2(s− t2) tan[2st(s− t2)]− 4s5t4tan[2st(s− t2)] + 2s4t
33. ∂u ∂s = ∂u ∂x ∂x ∂s + ∂u ∂y ∂y ∂s + ∂u ∂z ∂z ∂s
= (2x− y)(cos t) + (−x)(− cos (t − s)) + 2z(t cos s)
= 2s cos2t− sin (t − s) cos t + s cos t cos (t − s) + 2t2sin s cos s ∂u ∂t = ∂u ∂x ∂x ∂t + ∂u ∂y ∂y ∂t + ∂u ∂z ∂z ∂t
= (2x− y)(−s sin t) + (−x)(cos (t − s)) + 2z(sin s)
=−2s2cos t sin t + s sin (t− s) sin t − s cos t cos (t − s) + 2t sin2s
34. ∂u ∂s = ∂u ∂x· ∂x ∂s + ∂u ∂y · ∂y ∂s+ ∂u ∂z · ∂z ∂s = eyz21 s + xz 2eyz2 · 0 + 2xyzeyz2 2s = 1 se t3(s2+t2)2 + 4st3(s2+ t2) ln(st)et3(s2+t2)2 ∂u ∂t = ∂u ∂x· ∂x ∂t + ∂u ∂y · ∂y ∂t + ∂u ∂z · ∂z ∂t = eyz21 t + xz 2eyz2 3t2+ 2xyzeyz22t = 1 te t3(s2+t2)2 + t2(s2+ t2)(3s2+ 7t2) ln(st)et3(s2+t2)2 35. d dt[f (r(t) ) ] = ∇f(r(t) ) · rr(t) (t) r(t) = fu(t) (r(t)) r(t) where u(t) =
r(t) r(t) 36. ∂ ∂x[f (r)] = d dr[f (r)] ∂r ∂x = f (r)∂r ∂x = f (r)x r; similarly ∂ ∂y[f (r)] = f (r)y r and ∂ ∂z[f (r)] = f (r)z r. Therefore ∇f(r) = f(r)x ri + f (r)y rj + f (r)z rk = f (r)r r. 37. (a) (cos r)r r (b) (r cos r + sin r) r r 38. (a) ∇(r ln r) = (1 + ln r)r r (b) ∇(e 1−r2 ) =−2re1−r2r r =−2e1−r 2 r
39. (a) (r cos r− sin r) r r3 (b) sin r− r cos r sin2r r r 40. (a) (b) du dt = ∂u ∂x dx ds ds dt + ∂u ∂y dy ds ds dt 41. (a) (b) ∂u ∂r = ∂u ∂x ∂x ∂w ∂w ∂r + ∂x ∂t ∂t ∂r +∂u ∂y ∂y ∂w ∂w ∂r + ∂y ∂t ∂t ∂r +∂u ∂z ∂z ∂w ∂w ∂r + ∂z ∂t ∂t ∂r . To obtain ∂u/∂s, replace each r by s.
42. (a) (b) ∂u ∂r = ∂u ∂x ∂x ∂r + ∂u ∂z ∂z ∂r+ ∂u ∂w ∂w ∂r, ∂u ∂v = ∂u ∂y ∂y ∂v + ∂u ∂w ∂w ∂v
804
SECTION 16.3
43. du dt = ∂u ∂x dx dt + ∂u ∂y dy dt d2u dt2 = ∂u ∂x d2x dt2 + dx dt ∂2u ∂x2 dx dt + ∂2u ∂y∂x dy dt +∂u ∂y d2y dt2 + dy dt ∂2u ∂x ∂y dx dt + ∂2u ∂y2 dy dt and the result follows.44. ∂u ∂s = ∂u ∂x ∂x ∂s+ ∂u ∂y ∂y ∂s ∂2u ∂s2 = ∂u ∂x ∂2x ∂s2 + ∂x ∂s ∂2u ∂x2 ∂x ∂s + ∂2u ∂y∂x ∂y ∂s +∂u ∂y ∂2y ∂s2 + ∂y ∂s ∂2u ∂x∂y ∂x ∂s + ∂2u ∂y2 ∂y ∂s =∂ 2u ∂x2 ∂x ∂s 2 + 2 ∂ 2u ∂x∂y ∂x ∂s ∂y ∂s+ ∂2u ∂y2 ∂y ∂s 2 +∂u ∂x ∂2x ∂s2 + ∂u ∂y ∂2y ∂s2 45. (a) ∂u ∂r = ∂u ∂x ∂x ∂r + ∂u ∂y ∂y ∂r = ∂u ∂xcos θ + ∂u ∂y sin θ ∂u ∂θ = ∂u ∂x ∂x ∂θ + ∂u ∂y ∂y ∂θ = ∂u ∂x(−r sin θ) + ∂u ∂y(r cos θ) (b) ∂u ∂r 2 = ∂u ∂x 2 cos2θ + 2∂u ∂x ∂u
∂ycos θ sin θ + ∂u ∂y 2 sin2θ, 1 r2 ∂u ∂θ 2 = ∂u ∂x 2 sin2θ− 2∂u ∂x ∂u
∂ycos θ sin θ + ∂u ∂y 2 cos2θ, ∂u ∂r 2 + 1 r2 ∂u ∂θ 2 = ∂u ∂x 2 cos2θ + sin2θ+ ∂u ∂y 2 sin2θ + cos2θ= ∂u ∂x 2 + ∂u ∂y 2
46. (a) By Exercise 45 (a)
∂w ∂r = ∂w ∂x cos θ + ∂w ∂y sin θ, ∂w ∂θ =− ∂w ∂xr sin θ + ∂w ∂yr cos θ. Solve these equations simultaneously for ∂w
∂x and ∂w
∂y. (b) To obtain the first pair of equations set w = r;
to obtain the second pair of equations set w = θ. (c) θ is not independent of x; r =x2+ y2 gives
∂r ∂x = x x2+ y2 = r cos θ r = cos θ
47. Solve the equations in Exercise 45 (a) for ∂u
∂x and ∂u ∂y: ∂u ∂x = ∂u ∂r cos θ− 1 r ∂u ∂θ sin θ, ∂u ∂y = ∂u ∂r sin θ + 1 r ∂u ∂θ cos θ
Then ∇u = ∂u
∂xi + ∂u ∂yj = ∂u ∂r(cos θ i + sin θ j) + 1 r ∂u ∂θ(− sin θ i + cos θ j)
48. u(r, θ) = r2 =⇒ ∇u = 2re
49. u(x, y) = x2− xy + y2= r2− r2cos θ sin θ = r21−1 2 sin 2θ ∂u ∂r = r(2− sin 2θ), ∂u ∂θ =−r 2cos 2θ ∇u =∂u ∂rer+ 1 r ∂u
∂θeθ= r(2− sin 2θ)er− r cos 2θ eθ
50. ∂u ∂θ = ∂u ∂x ∂x ∂θ + ∂u ∂y ∂y ∂θ =−r sin θ ∂u ∂x + r cos θ ∂u ∂y ∂2u ∂r∂θ =− sin θ ∂u ∂x − r sin θ ∂2u ∂x2 ∂x ∂r + ∂2u ∂y∂x ∂y ∂r + cos θ∂u ∂y + r cos θ ∂2u ∂x∂y ∂x ∂r + ∂2u ∂y2 ∂y ∂r =− sin θ∂u ∂x + cos θ ∂u
∂y + r sin θ cos θ ∂2u ∂y2 − ∂2u ∂x2 + r(cos2θ− sin2θ) ∂ 2u ∂x∂y
51. From Exercise 45 (a),
∂2u
∂r2 =
∂2u
∂x2 cos
2θ + 2 ∂2u
∂y ∂x sin θ cos θ + ∂2u ∂y2 sin 2θ ∂2u ∂θ2 = ∂2u ∂x2r 2sin2θ− 2 ∂2u ∂y ∂xr 2sin θ cos θ +∂2u ∂y2 r 2cos2θ− r ∂u ∂xcos θ + ∂u ∂y sin θ . The term in parentheses is ∂u
∂r. Now divide the second equation by r
2 and add the two equations.
The result follows.
52. u(x, y) = x2− 2xy + y4− 4, ∂u
∂x = 2x− 2y, ∂u ∂y =−2x + 4y 3 dy dx =− ∂u ∂x ∂u ∂y = 2y− 2x 4y3− 2x = y− x 2y3− x 53. Set u = xey+ yex− 2x2y. Then ∂u ∂x = e y+ yex− 4xy, ∂u ∂y = xe y+ ex− 2x2 dy dx =− ∂u/∂x ∂u/∂y =− ey+ yex− 4xy xey+ ex− 2x2. 54. u(x, y) = x2/3+ y2/3, ∂u ∂x = 2 3x −1/3, ∂u ∂y = 2 3y −1/3 =⇒ dy dx =− ∂u ∂x ∂u ∂y =− y x 1/3
55. Set u = x cos xy + y cos x− 2. Then ∂u
∂x = cos xy− xy sin xy − y sin x,
∂u ∂y =− x 2sin xy + cos x dy dx =− ∂u/∂x ∂u/∂y =
cos xy− xy sin xy − y sin x x2sin xy− cos x .
806
SECTION 16.4
56. Set u(x, y, z) = z4+ x2z3+ y2+ xy− 2. Then ∂u
∂x = 2xz 3+ y, ∂u ∂y = 2y + x, ∂u ∂z = 4z 3+ 3x2z2 ∂z ∂x =− ∂u ∂x ∂u ∂z =− 2xz 3+ y 4z3+ 3x2z2, ∂z ∂y =− ∂u ∂y ∂u ∂z =− 2y + x 4z3+ 3x2z2
57. Set u = cos xyz + lnx2+ y2+ z2. Then
∂u
∂x =−yz sin xyz +
2x x2+ y2+ z2,
∂u
∂y =−xz sin xyz +
2y
x2+ y2+ z2, and
∂u
∂z =−xy sin xyz +
2z x2+ y2+ z2. ∂z ∂x =− ∂u/∂x ∂u/∂z =− 2x− yzx2+ y2+ z2sin xyz 2z− xy (x2+ y2+ z2) sin xyz, ∂z ∂y =− ∂u/∂y ∂u/∂z =− 2y− xzx2+ y2+ z2sin xyz 2z− xy (x2+ y2+ z2) sin xyz. 58. (a) Use du dt = du1 dt i + du2
dt j and apply the chain rule to u1, u2. (b) (i) du
dt = t(e
xcos yi + exsin yj) + π(−exsin yi + excos yj) = tet2/2(cos πti + sin πtj) + πet2/2(− sin πti + cos πtj) (ii) u(t) = et2/2cos πti + et2/2sin πtj
du
dt = (−πe t2/2
sin πt + tet2/2cos πt)i + (πet2/2cos πt + tet2/2sin πt)j
59. ∂u ∂s = ∂u ∂x ∂x ∂s + ∂u ∂y ∂y ∂s, ∂u ∂t = ∂u ∂x ∂x ∂t + ∂u ∂y ∂y ∂t 60. du dt = ∂u ∂x dx dt + ∂u ∂y dy dt + ∂u ∂z dz dt where ∂u ∂x = ∂u1 ∂x i + ∂u2 ∂x j + ∂u3 ∂x k, ∂u ∂y = ∂u1 ∂y i + ∂u2 ∂y j + ∂u3 ∂y k, ∂u ∂z = ∂u1 ∂z i + ∂u2 ∂z j + ∂u3 ∂z k. SECTION 16.4 1. Set f (x, y) = x2+ xy + y2. Then,
∇f = (2x + y)i + (x + 2y)j, ∇f(−1, −1) = −3i − 3j.
normal vector i + j; tangent vector i− j
2. Set f (x, y) = (y− x)2− 2x, ∇f = −2(y − x + 1)i + 2(y − x)j, ∇f(2, 4) = −6i + 4j
normal vector −3i + 2j; tangent vector 2i + 3j
tangent line 3x− 2y + 2 = 0; normal line 2x + 3y − 16 = 0
3. Set f (x, y) =x2+ y22− 9x2− y2. Then,
∇f = [4x(x2+ y2)− 18x]i +4yx2+ y2+ 18yj, ∇f √2, 1=−6√2 i + 30j.
normal vector√2 i− 5 j; tangent vector 5i +√2 j
tangent line√2x− 5y + 3 = 0; normal line 5x +√2 y− 6√2 = 0
4. Set f (x, y) = x3+ y3, ∇f = 3x2i + 3y2j, ∇f(1, 2) = 3i + 12j
normal vector i + 4j; tangent vector 4i− j
tangent line x + 4y− 9 = 0; normal line 4x− y − 2 = 0
5. Set f (x, y) = xy2− 2x2+ y + 5x. Then,
∇f = (y2− 4x + 5) i + (2xy + 1) j, ∇f(4, 2) = −7i + 17j.
normal vector 7i− 17j; tangent vector 17i + 7j
tangent line 7x− 17y + 6 = 0; normal line 17x + 7y − 82 = 0
6. Set f (x, y) = x5+ y5− 2x3. ∇f = (5x4− 6x2)i + 5y4j, ∇f(1, 1) = −i + 5j
normal vector i− 5j; tangent vector 5i + j
tangent line x− 5y + 4 = 0; normal line 5x + y− 6 = 0
7. Set f (x, y) = 2x3− x2y2− 3x + y. Then,
∇f = (6x2− 2xy2− 3) i + (−2x2y + 1) j, ∇f(1, −2) = −5i + 5j.
normal vector i− j; tangent vector i + j
tangent line x− y − 3 = 0; normal line x + y + 1 = 0
8. Set f (x, y) = x3+ y2+ 2x. ∇f = (3x2+ 2)i + 2yj, ∇f(−1, 3) = 5i + 6j
normal vector 5i + 6j; tangent vector 6i− 5j
tangent line 5x + 6y− 13 = 0; normal line 6x− 5y + 21 = 0
9. Set f (x, y) = x2y + a2y. By (15.4.4) m =−∂f /∂x ∂f /∂y =− 2xy x2+ a2. At (0, a) the slope is 0. 10. Set f (x, y, z) = (x2+ y2)2− z. ∇f = 4x(x2+ y2)i + 4y(x2+ y2)j− k, ∇f(1, 1, 4) = 8i + 8j − k Tangent plane: 8x + 8y− z − 12 = 0 Normal: x = 1 + 8t, y = 1 + 8t, z = 4− t
808
SECTION 16.4
11. Set f (x, y, z) = x3+ y3− 3xyz. Then,
∇f = (3x2− 3yz) i + (3y2− 3xz) j − 3xyk, ∇f1, 2,3 2
=−6i +152j− 6k;
tangent plane at1, 2,32: −6(x − 1) +152(y− 2) − 6z−32= 0, which reduces to 4x− 5y + 4z = 0. Normal: x = 1 + 4t, y = 2− 5t, z = 32+ 4t
12. Set f (x, y, z) = xy2+ 2z2. ∇f = y2i + 2xyj + 4zk, ∇f(1, 2, 2) = 4i + 4j + 8k
Tangent plane: x + y + 2z− 7 = 0
Normal: x = 1 + t, y = 2 + t, z = 2 + 2t
13. Set z = g(x, y) = axy. Then, ∇g = ayi + axj, ∇g 1,1 a = i + aj. tangent plane at 1,1 a, 1 : z− 1 = 1(x − 1) + a y−1 a , which reduces to x + ay− z − 1 = 0 Normal: x = 1 + t, y = 1a+ at, z = 1− t 14. Set f (x, y, z) =√x +√y +√z. ∇f = 1 2√xi + 1 2√yj + 1 2√zk, ∇f(1, 4, 1) = 1 2i + 1 4j + 1 2k Tangent plane: 2x + y + 2z− 8 = 0 Normal: x = 1 + 2t, y = 4 + t, z = 1 + 2t
15. Set z = g(x, y) = sin x + sin y + sin (x + y). Then,
∇g = [cos x + cos (x + y)] i + [cos y + cos (x + y)] j, ∇g(0, 0) = 2i + 2j;
tangent plane at (0, 0, 0) : z− 0 = 2(x − 0) + 2(y − 0), 2x + 2y − z = 0. Normal: x = 2t, y = 2t, z =−t
16. Set f (x, y, z) = x2+ xy + y2− 6x + 2 − z. ∇f = (2x + y − 6)i + (x + 2y)j − k, ∇f(4, −2, −10) = −k Tangent plane: z =−10 Normal: x = 4, y =−2, z = −10 + t 17. Set f (x, y, z) = b2c2x2− a2c2y2− a2b2z2. Then, ∇f (x0, y0, z0) = 2b2c2x0i− 2a2c2y0j− 2a2b2z0k; tangent plane at (x0, y0, z0) : 2b2c2x0(x− x0)− 2a2c2y0(y− y0)− 2a2b2z0(z− z0) = 0,
which can be rewritten as follows:
b2c2x0x− a2c2y0y− a2b2z0z = b2c2x02− a2c2y02− a2b2z02
= f (x0, y0, z0) = a2b2c2.
18. Set f (x, y, z) = sin(x cos y)− z. ∇f = cos y cos(x cos y)i − x sin y cos(x cos y)j − k, ∇f(1,π 2, 0) = j− k Tangent plane: y + z =π 2 Normal: x = 1, y = π 2 + t, z = t 19. Set z = g(x, y) = xy + a3x−1+ b3y−1. ∇g =y− a3x−2i +x− b3y−2j, ∇g = 0 =⇒ y = a3x−2 and x = b3y−2. Thus,
y = a3b−6y4, y3= b6a−3, y = b2/a, x = b3y−2= a2/b and ga2/b, b2/a= 3ab.
The tangent plane is horizontal ata2/b, b2/a, 3ab.
20. z = g(x, y) = 4x + 2y− x2+ xy− y2. ∇g = (4 − 2x + y)i + (2 + x − 2y)j
∇g = 0 =⇒ 4 − 2x + y = 0, 2 + x − 2y = 0 =⇒ x =10
3 , y = 8 3 The tangent plane is horizontal at (103,83,283).
21. Set z = g(x, y) = xy. Then, ∇g = yi + xj.
∇g = 0 =⇒ x = y = 0.
The tangent plane is horizontal at (0, 0, 0).
22. z = g(x, y) = x2+ y2− x − y − xy. ∇g = (2x − 1 − y)i + (2y − 1 − x)j
∇g = 0 =⇒ 2x − 1 − y = 0 = 2y − 1 − x = 0 =⇒ x = 1, y = 1
The tangent plane is horizontal at (1, 1,−1).
23. Set z = g(x, y) = 2x2+ 2xy− y2− 5x + 3y − 2. Then,
∇g = (4x + 2y − 5) i + (2x − 2y + 3) j. ∇g = 0 =⇒ 4x + 2y − 5 = 0 = 2x − 2y + 3 =⇒ x = 1
3, y = 11
6.
The tangent plane is horizontal at13, 116,−121.
24. (a) Set f (x, y, z) = xy− z. ∇f = yi + xj − k, ∇f(1, 1, 1) = i + j − k
upper unit normal = √ 3 3 (−i − j + k) (b) Set f (x, y, z) = 1 x− 1 y − z. ∇f = − 1 x2i + 1 y2j− k, ∇f(1, 1, 0) = −i + j − k
lower unit normal: = √ 3 3 (−i + j − k) 25. x− x0 (∂f /∂x)(x0, y0, z0) = y− y0 (∂f /∂y)(x0, y0, z0) = z− z0 (∂f /∂z)(x0, y0, z0)
810
SECTION 16.4
26. All the tangent planes pass through the origin. To see this, write the equation of the surface as
xf (x/y)− z = 0. The tangent plane at (x0, y0, z0) has equation
(x− x0) x0 y0 f x0 y0 + f x0 y0 − (y − y0) x02 y02 f x0 y0 − (z − z0) = 0.
The plane passes through the origin: −x02 y0 f x0 y0 − x0f x0 y0 +x0 2 y0 f x0 y0 + z0= z0− x0f x0 y0 = 0.
27. Since the tangent planes meet at right angles, the normals∇F and ∇G meet at right angles:
∂F ∂x ∂G ∂x + ∂F ∂y ∂G ∂y + ∂F ∂z ∂G ∂z = 0.
28. The sum of the intercepts is a. To see this, note that the equation of the tangent plane at
(x0, y0, z0) can be written x√− x0 x0 +y√− yy 0 0 +z√− z0 z0 = 0. Setting y = z = 0 we have x− x0 √x 0 =√y0+√z0.
Therefore the x-intercept is given by
x = x0+√x0(√y0+√z0) = x0+√x0(
√
a−√x0) =√x0
√ a.
Similarly the y-intercept is√y0√a and the z-intercept is√z0√a. The sum of the intercepts is
(√x0+√y0+√z0)
√
a =√a√a = a.
29. The tangent plane at an arbitrary point (x0, y0, z0) has equation
y0z0(x− x0) + x0z0(y− y0) + x0y0(z− z0) = 0, which simplifies to y0z0x + x0z0y + x0y0z = 3x0y0z0 and thus to x 3x0 + y 3y0 + z 3z0 = 1. The volume of the pyramid is
V = 1 3Bh = 1 3 (3x0) (3y0) 2 (3z0) = 9 2x0y0z0= 9 2a 3.
30. The equation of the tangent plane at (x0, y0, z0) can be written
x0−1/3(x− x0) + y0−1/3(y− y0) + z0−1/3(z− z0) = 0
Setting y = z = 0, we get the x-intercept x = x0+ x01/3(y02/3+ z02/3) = x0+ x01/3(a2/3− x02/3)
=⇒ x = x01/3a2/3
Similarly, the y-intercept is y01/3a2/3 and the z-intercept is z01/3a2/3.
The sum of the squares of the intercepts is
